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JEE PHYSICS MCQ: THERMAL PROPERTIES OF MATTER

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A normal diet furnishes 2000 kcal2000 \mathrm{~kcal} to a 60 kg60 \mathrm{~kg} person in a day. If this energy was used to heat the person with no losses to the surroundings, how much would the person's temperature increase? The specific heat of the human body =0.83 cal g−1∘C−1=0.83 \mathrm{~cal} \mathrm{~g}^{-1}{ }^{\circ} \mathrm{C}^{-1}.
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Correct option: A
40.16∘C40.16^{\circ} \mathrm{C}
Option A (40.16∘C40.16^{\circ} \mathrm{C}) is correct because it is calculated using the formula ΔT=Qmc\Delta T = \frac{Q}{mc}, where Q is the heat energy, m is the mass, and c is the specific heat.

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